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Codeforces Round #336 (Div. 2) B 暴力

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B. Hamming Distance Sum time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output Genos needs your help. He was asked to solve the following programming problem by Saitama:

The length of some string s is denoted |s|. The Hamming distance between two strings s and t of equal length is defined as

Codeforces Round #336 (Div. 2) B 暴力

, where si is the i-th character of s and ti is the i-th character of t. For example, the Hamming distance between string "0011" and string "0110" is |0 - 0| + |0 - 1| + |1 - 1| + |1 - 0| = 0 + 1 + 0 + 1 = 2.

Given two binary strings a and b, find the sum of the Hamming distances between a and all contiguous substrings of b of length |a|.

Input

The first line of the input contains binary string a (1 ≤ |a| ≤ 200 000).

The second line of the input contains binary string b (|a| ≤ |b| ≤ 200 000).

Both strings are guaranteed to consist of characters '0' and '1' only.

Output

Print a single integer — the sum of Hamming distances between a and all contiguous substrings of b of length |a|.

Examples

input

01

00111

output

3

input

0011

0110

output

2

Note

For the first sample case, there are four contiguous substrings of b of length |a|: "00", "01", "11", and "11". The distance between "01" and "00" is |0 - 0| + |1 - 0| = 1. The distance between "01" and "01" is |0 - 0| + |1 - 1| = 0. The distance between "01" and "11" is |0 - 1| + |1 - 1| = 1. Last distance counts twice, as there are two occurrences of string "11". The sum of these edit distances is 1 + 0 + 1 + 1 = 3.

The second sample case is described in the statement.

题意:

给出两串01串s,p 要求p中所有长度为len(s)的串与s串相减绝对值,然后累加起来并输出

思路:

嗯 维护一下前缀和然后算贡献

代码:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 1000100
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
string s,p;
int sum[200100];
int a[200100];
int b[200100];
int main(){
    cin>>s>>p;
    int n=s.size(),m=p.size();
    for(int i=1;i<=n;i++){
        b[i]=s[i-1]-'0';
        b[i]*=(m-n+1);
    }
    sum[0]=0;
    for(int i=1;i<=m;i++){
        int a=p[i-1]-'0';
        sum[i]=sum[i-1]+a;
    }
    int cnt=0;
    for(int i=m-n+1;i<=m;i++){
        a[++cnt]=sum[i]-sum[i-(m-n+1)];
    }
    ll ans=0;
    for(int i=1;i<=n;i++){
        ans+=abs(b[i]-a[i]);
    }
    cout<<ans<<endl;
    return 0;
}
/*
1001101001101110101101000
01111000010011111111110010001101000100011110101111
*/
           

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